Let $\mathbb F$ be an extension of a field $\mathbb K$. Let $\alpha \in \mathbb F$ be an element of odd degree over $\mathbb K$. Show that $\alpha^2$ is also an odd degree algebraic over $\mathbb K$, and $\mathbb K(\alpha) = \mathbb K (\alpha^2)$.
$F(\alpha)$ is a finite extension of $F$ and $\alpha^2 \in F(\alpha)$, so $\alpha^2$ is algebraic over $F$. If $F(\alpha^2) \neq F(\alpha)$, then $F(\alpha)$ must be an extension of $F(\alpha^2)$ of degree 2, because is a zero of $x^2 − \alpha^2$. Then we would have $2 = [F(\alpha) : F(\alpha^2)]$ dividing $[F(\alpha) : F]$, which is impossible because $[F(\alpha) : F]$ is an odd number. Therefore $F(\alpha^2) = F(\alpha)$, so deg(irr$(\alpha^2, F)$) = deg(irr($\alpha, F$)) = $[F(\alpha) : F]$ which is an odd number.
That's what I managed to do.
Opinions?
Thank's for any help.
Best Answer
You are in the correct path, the argument is justo that the extension $K(\alpha)/K(\alpha^2)$ has degree at most two, since it's obtained by adjoining $\alpha$, a root of a quadratic equation, to $K(\alpha^2)$. The quadratic is $X^2-\alpha^2=0$. But its degree is odd, so is one, that is $K(\alpha)=K(\alpha^2)$.